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3 years ago

The sum of two numbers is 10. twice the first number is 4 more than twice the second. what are the two numbers?

1 answer:

6 0

X+ y = 10
2x = 2y+4

2x/2 = 2/2y+4/2
x = y + 2

x+y = 10
y+2+y=10
2y + 2 = 10
2y +2-2 =10-2
2y = 8
2y/2 =8/2
y = 4

x + y =10
x + 4 =10
x +4-4 =10-4
x = 6

Check
2x = 2y +4
2(6) = 2(4) +4
12 = 8+4
12= 12

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Determine the quotient, q(x), and remainder, r(x) when

Vinvika [58]

ANSWER

Quotient:

$q(x) = 4 {x}^{2} + 3x + 2$

Remainder:

$r(x) = 3x + 1$

EXPLANATION

The given functions are

$f(x) = 12 {x}^{4} + 21 {x}^{3} + 31 {x}^{2} + 21x + 9$

and

$g(x) = 3 {x}^{2} + 3x + 4$

We want to find the remainder and quotient when f(x) is divided by g(x).

We perform the long division as shown in the attachment.

The quotient is

$q(x) = 4 {x}^{2} + 3x + 2$

The remainder is

$r(x) = 3x + 1$

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3 years ago

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

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2 years ago

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